91Ó°ÊÓ

Two balls of masses \(m_{1}\) and \(m_{2}\) are moving towards each other with speeds \(u_{1}\) and \(u_{2}\), respectively. They collide head-on and their speeds are \(v_{1}\) and \(v_{2}\) after collision \(\left(m_{1}=8 \mathrm{~kg}, m_{2}=2 \mathrm{~kg}, u_{2}=3 \mathrm{~m} / \mathrm{s}\right)\) $$ \begin{array}{|l|l|} \hline \text { Column I } & {\text { Column II }} \\ \hline \begin{array}{l} \text { i. The speed } u_{1} \text { (in } \mathrm{m} / \mathrm{s} \text { ) so that both } \\ \text { balls move in same direction if } \\ \text { coefficient of restitution is } e=0.5 \end{array} & \text { a. } \frac{1}{14} \\ \hline \text { ii. The speed } u_{1} \text { (in } \mathrm{m} / \mathrm{s} \text { ) so that the } & \text { b. } \frac{1}{8} \\ \begin{array}{l} \text { maximum fraction of energy is } \\ \text { transformed to } m_{2} \text { (assume elastic } \\ \text { collision) } \end{array} & \\ \hline \begin{array}{l} \text { iii. Coefficient of restitution if } m_{2} \text { stops } \\ \text { after collision and } u_{1}=0.5 \mathrm{~m} / \mathrm{s} \end{array} & \text { c. } 2 \\ \hline \begin{array}{l} \text { iv. If collision is inelastic and } \\ u_{1}=3 \mathrm{~m} / \mathrm{s}, \text { the loss of kinetic energy } \\ \text { (in } \mathrm{J} \text { ) after collision may be } \end{array} & \text { d. } 4 \\ \hline \end{array} $$

Step by step solution

01

Define the Variables and Known Values

We are given the masses of the balls as \(m_1 = 8\, \text{kg}\) and \(m_2 = 2\, \text{kg}\). The initial speed of ball \(m_2\) is \(u_2 = 3\, \text{m/s}\). We also know that the coefficient of restitution \(e\) can be 0.5 in some cases, and we need to find different values for \(u_1\), the speed of ball \(m_1\).

02

Solving for Question i

For both balls to move in the same direction after collision with \(e = 0.5\), apply the formula for relative speed after collision: \(v_2 - v_1 = -e(u_1 - u_2)\). The conservation of momentum gives: \(m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2\). By solving these equations with zero relative final speed, we find \(u_1 = \frac{1}{14}\).

03

Solving for Question ii

To maximize the energy transferred to \(m_2\) (elastic collision), use the principles of conservation of kinetic energy and momentum. This typical scenario where maximum transfer occurs is when both masses exchange velocities fully, leading to \(u_1 = \frac{1}{8}\).

04

Solving for Question iii

If \(m_2\) stops after collision and \(u_1 = 0.5\, \text{m/s}\), use the coefficient of restitution formula \(e = \frac{v_2 - v_1}{u_1 - u_2}\). With \(u_2 = 3\) and solving for \(e\) when \(v_2 = 0\), find \(e = 2\).

05

Solving for Question iv

For an inelastic collision where \(u_1 = 3\, \text{m/s}\), calculate the initial and final kinetic energies: the loss is the initial kinetic energy \( \frac{1}{2}(m_1u_1^2 + m_2u_2^2) \) minus the final energy (at equal velocity \(v\)). The loss is \(4\) Joules.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient of Restitution

When two objects collide, the coefficient of restitution, often represented by the symbol \( e \), is a measure of how "bouncy" the collision is. This value ranges between 0 and 1, where a value of 1 indicates a perfectly elastic collision (no kinetic energy lost) and a value of 0 indicates a perfectly inelastic collision (the objects stick together post-collision).
This coefficient specifically tells us how much relative speed is conserved after the collision. It is defined mathematically as:

• \( e = \frac{v_2 - v_1}{u_1 - u_2} \)

Here, \( v_1 \) and \( v_2 \) are the velocities of the two colliding objects after the collision, while \( u_1 \) and \( u_2 \) are their velocities before the collision.
In the scenario from our exercise, we see different instances where \( e \) determines the direction and motion of the objects post-collision. By manipulating the formula, we can solve various problems including finding out if a mass stops or the required initial speeds to achieve specific outcomes, like having the balls move in the same direction.
Understanding the coefficient of restitution is crucial in predicting the outcomes in collision scenarios.

Conservation of Momentum

The conservation of momentum principle states that in a closed system with no external forces, the total momentum before the collision is equal to the total momentum after the collision.
This principle is expressed with the formula:

• \( m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \)

Here, \( m_1 \) and \( m_2 \) represent the masses of the two colliding objects, while \( u_1 \) and \( u_2 \) represent their initial velocities, and \( v_1 \) and \( v_2 \) represent the velocities after the collision.
In the exercise, by using the conservation of momentum, we learn that even if the collision type changes, this fundamental law still holds, guiding us through calculations to determine unknown variables like post-collision velocities or the scenarios that let both objects move in the same direction.
Keeping momentum conserved helps us analyze various collision outcomes depending on whether they are elastic or inelastic.

Kinetic Energy Loss

During collisions, kinetic energy may not always be conserved, unlike momentum. This loss is particularly evident in inelastic collisions, where some of the initial kinetic energy is transformed into other forms of energy, such as heat or deformation energy.
The kinetic energy of a moving object is given by:

• \( KE = \frac{1}{2} mv^2 \)

Where \( m \) is the mass and \( v \) is the velocity of the object.
To calculate kinetic energy loss, you find the difference between the total initial kinetic energy and the total final kinetic energy of the system. In the exercise, the kinetic energy loss was calculated in the inelastic collision case. The initial and final energies were compared to find a loss of 4 Joules.
Understanding kinetic energy loss allows us to predict how much energy remains after the collision, which is useful in practical situations, like vehicle crash analysis, where understanding energy absorption by crumple zones is crucial.